Finding Equilibrium in a Hanging Beam System

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A 6.00-meter uniform beam is suspended at an angle from a point right of its center, weighing 138 kg. A concrete block is attached at one end, while an unknown weight is at the other end, and the system is in equilibrium. Participants seek clarification on the weight of the concrete block and the angle of the beam with the vertical to solve for the unknown weight. The discussion emphasizes the need for specific values to apply the equilibrium equations effectively. Understanding these parameters is crucial for determining the unknown weight in the hanging beam system.
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Homework Statement



In the figure a 6.00-m-long, uniform beam is hanging from a point to the right of its center. The beam weighs 138 and makes an angle of with the vertical. At the right-hand end of the beam a concrete block weighing is hung; an unknown weight hangs at the other end.
If the system is in equilibrium, what is w ? You can ignore the thickness of the beam.

Homework Equations



F=ma

The Attempt at a Solution



i don't rly know where to start i just need a hint
 
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The beam weighs 138 and makes an angle of with the vertical

Does the beam weigh 138 kg? What is the angle that is made with the vertical?

At the right-hand end of the beam a concrete block weighing is hung

How much does this block weigh?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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